Non-conductve sphere with cavity -- find Electric field

[Manolisjam]

I have a non conducting sphere with a charge ρ=A/r per uni vollume A is constant. suppose there is a cavity in the centre and within a particle of charge q. i want to find the E inside the sphere in respect with r.

Homework Equations

The Attempt at a Solution

for radius equal of the cavity i get $E=kq/r^2$ for r bigger than the radius of the sphere $E=k(Q+q)/r^2$ now only one case left
if r is bigger than the radius of the cavity and smaller than the radius of the sphere.
$E4πr^2=\frac{q+ρV}{ε_0}$
where ρ is the density per unit vollume and V is the vollume
is my enclosed Q right? and my V is 4π(r^3-a^3) where a^3 is the small radius.

 

[gneill]

is my enclosed Q right?

Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.

 

[Manolisjam]

Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.

?

 

[gneill]

Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.

 

[Manolisjam]

Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.

$E4πr^2=\frac{q+ \int ρVdr}{ε_0}$ is the right one?

 

[gneill]

$E4πr^2=\frac{q+ \int ρVdr}{ε_0}$ is the right one?

That's the idea, yes.

 

[Manolisjam]

That's the idea, yes.

so i Keep the vollume inside the integral!

 

[gneill]

You want to integrate over the volume of interest, summing up the charges contained in each differential volume element. So really that V should be a dV(r). I assumed (my bad) that that was what you were implying with that V.